Question

Get the formula $△=\sqrt{s(s-a)(s-b)(s-c)}$ for the area of a triangle by use of vectors.

Answer 1

Assume the Triangle has vertices $O,A,B$

$\left||A\right||=a,\left||B\right||=b,||B-A||=c$

$s=\frac{a+b+c}{2}$

Consider $S=\frac{1}{2}\left||A\times B|\right|$

$4S^2={\left|\left|A\right|\right|}^2{\left|\left|B\right|\right|}^2-(A.B)$

$⟹4S^2=a^2{b}^2-\frac{1}{4}(c^2-a^2-b^2{)}^2=\frac{1}{4}(2ab-c^2+a^2+b^2\left)\right(2ab+c^2-a^2-b^2)$

$⟹S^2=\frac{1}{16}(a+b+c)(a+b-c)(c-a+b)(c+a-b)$

$⟹S=\sqrt{s(s-a)(s-b)(s-c)}$